For this one, I think you have to use the pythagorean thm to relate the diameter to the breadth (b) & height (h). The diameter is d = sqrt{b^2 + h^2} so d^2 = b^2 + h^2 = 12^2 = 144. Solve this for b to get b = sqrt{144 - h^2} If stiffness is S then S = bh^3 = sqrt{144 - h^2}*h^3. Do the derivative to find the max/min.

Thank you for putting me on the right track.Solving the derivative of: (144-h^2)^0.5 * h^3 = 0, this gives: 3 h^2(144-h^2)^0.5 - h^4(144-h^2)^-0.5 = 0 (by the poduct rule).Solving this eqn: h = -6*sqrt(3) and h = 6*sqrt(3), (only the positive value is valid).So the height should be 10.4 inches and the width 6 inches for max. stiffness of this beam.S = bh^3 gives a maximum value for these results, so I assume this must be the right answer.Thanks again.Luke.